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Let $\mathcal{P}^n$ denote the vector space of homogeneous polynomials on $\mathbb{R}^3$ of degree $n$. I need to prove that $\Delta|_{\mathcal{P}^n}:\mathcal{P}_n\to\mathcal{P}_{n-2}$, for $n\geq2$ is surjective, where $\Delta$ is the Laplace operator.

The hint says that the proof should be done by inductive argument: in the inductive step I should conclude from the formula $$\Delta(x_1^{q_1}x_2^{q_2}x_3^{q_3})=q_1(q_1-1)x_1^{q_1-2}x_2^{q_2}x_3^{q_3}+q_2(q_2-1)x_1^{q_1}x_2^{q_2-2}x_3^{q_3}+q_3(q_3-1)x_1^{q_1}x_2^{q_2}x_3^{q_3-2}$$ that surjectivity of $\Delta|_{\mathcal{P}^n}:\mathcal{P}_n\to\mathcal{P}_{n-2}$ implies surjectivity of $\Delta|_{\mathcal{P}^{n+2}}:\mathcal{P}_{n+2}\to\mathcal{P}_n$.

I've tried things and things and I simply don't see how to do this. Help!

sonjcy
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1 Answers1

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This was a neat problem, thanks for asking. Since it's homework, I'm going to skimp on some details. First check that: $$\Delta(pq) = q \Delta p + p \Delta q + 2\sum_i \partial_i p \partial_i q$$

Let's show that $\Delta: \mathcal{P}^{n+2} \rightarrow \mathcal{P}^{n}$ is a surjection. It's manifestly linear, so it suffices to see each monomial $x^iy^jz^k$ is in the image, for $i+j+k = n$. Let's strongly induct downwards on $\max\{i,j,k\}$. The base case of $n$ is clear. Suppose we know we can get all of them with $\max\{i,j,k\} \geqslant M$, let's get $M-1$.

WLOG we want to obtain the monomial $x^{M-1} y^- z^-$ (i.e. $x$ may as well be the variable with the largest power). Well, $\Delta$ is sorta like twice differentiating, so let's try $\Delta(x^{M+1} y^- z^-)$. From the above product rule, with $p = x^{M+1}, q = y^-z^-$, and ignoring constants, we have this equals $$\Delta(x^{M+1} y^- z^-) = x^{M-1} y^- z^- + x^{M+1}\Delta(y^- z^-) $$(check that $2 \sum_i \partial_i p \partial_i q$ vanishes). By inductive hypothesis, the second term lies in the image of $\Delta$ ($x$ appears to a high enough power), hence so does $\Delta(x^{M+1} y^- z^-) - x^{M+1} \Delta(y^- z^-)$, as desired.

uncookedfalcon
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  • Wow! I haven't thought of doing it this way - I was trying to listen to the hint and to connect the case for $\mathcal{P}n$ with $\mathcal{P}{n-2}$ and I was stuck, but now I see that it can be done (much simpler, it seems) without that. Ah, hints... Thank you veeeery much! :) – sonjcy Jan 23 '13 at 07:49
  • You're most welcome! – uncookedfalcon Jan 23 '13 at 20:40